Why Does ' []{}' Enable Lambda Expression Compilation in C ?

Unraveling the Mystery of ' ' with Lambda Expressions: ' []{}'

In a previous Stack Overflow question, a seemingly simple lambda expression failed to compile. However, with the addition of a ' ' operator before the lambda, the code miraculously gained its ability to compile. This intriguing observation poses the question: why does ' []{}' work?

The key lies in the ' ' operator's overloaded behavior. When applied to a closure object generated by a non-capturing lambda, it invokes a built-in conversion function that transforms the closure into a plain function pointer.

This conversion is crucial because the ' ' operator has a candidate overload that converts any type to a pointer. Therefore, applying ' ' to the closure object results in a function pointer to the lambda.

The type of 'test' after the first lambda is declared becomes void(*)(), which is a function pointer with a void return type and no parameters. This allows the second lambda to be assigned to 'test', despite having a different closure type.

Hence, the code's functionality can be explained as follows:

1. The first lambda is converted to a function pointer using the ' ' operator.
2. The type of 'test' becomes a function pointer void(*)().
3. The second lambda is also converted to a function pointer, ensuring compatibility with 'test'.
4. The assignment to 'test' succeeds because both function pointers are compatible.

This behavior is fully compliant with the C standard, making ' []{}' a valid and surprisingly useful trick for lambda expressions.

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